we have $\displaystyle z=1+3x+2y^2$ and I am to find the part of the surface that lies above a triangle with vertices at (0,0),(0,1),(2,1).

So far:

$\displaystyle dS=\sqrt{1+(\frac{dz}{dx})^2+(\frac{dz}{dy})^2}dA= \sqrt{1+(3)^2+(4y)^2}dA=\sqrt{16y^2+10}dA$, so the integral looks ugly...